Page 120 - Fundamentals of Communications Systems

3.34 Chapter Three

(b)
π
E[ p ] = xf (x)dx = 0 (3.49)
−π

Since p is a zero mean random variable

π 1 π 3 2
2 2 2 π π
var( p ) = x f (x)dx = x dx = = (3.50)
−π 2π −π 3 2π 3
(c)
1
◦ ◦ ◦ ◦
P (Work) = P (−30 ≤ p ≤ 30 ) = F p (30 ) − F p (−30 ) = (3.51)
6
(d) Poincaire’s theorem gives

P (Work 1 or Work 2 ) = P W 1 ∪ W 2 ) = P (W 1 ) + P (W 2 ) − P (W 1 ∩ W 2 ) (3.52)
Because of independence of the two trials

P (W 1 ∩ W 2 ) = P (W 1 )P (W 2 ) (3.53)
consequently

1 1 1 11
P (W 1 ∪ W 2 ) = P (W 1 ) + P (W 2 ) − P (W 1 )P (W 2 ) = + − = (3.54)
6 6 36 36
(e) Getting your receiver to work once in N trials is the complement to the
event not getting your receiver to work any of N trials, i.e.,
N N
C
P W i = 1 − P W (3.55)
i
i=1 i=1
Again by independence

N N
C C N
P W ) = P (W = (1 − P (W i )) (3.56)
i i
i=1 i=1
so that
N
N
P W i = 1 − (1 − P (W i )) (3.57)
i=1
A little exploring in Matlab shows that N ≥ 13 will give better than a 90%
probability of working.